I Olympiad Problem -- Revisiting the problem with 3 blocks and a pulley

[PeroK]

I assumed that (in the rest frame of M) the sum of gravity and inertial force on the hanging m is parallel to the string. This lead me to the following relationship:
$$\mu= \frac{1}{2} \: cos(\theta) \: cot(\theta)$$

I get $$\mu= \frac{1}{2} (1- \sin(\theta)) \cot(\theta)$$I got that two different ways. Initially by an analysis of forces looking for a constant angle. Then, by using the E-L equations with constant angle.

I think it makes more sense that when μ goes to 0, then θ approaches π/2

The full formula above does this. But not, of course, the small angle approximation.

 

[A.T.]

I get $$\mu= \frac{1}{2} (1- \sin(\theta)) \cot(\theta)$$I got that two different ways. Initially by an analysis of forces looking for a constant angle. Then, by using the E-L equations with constant angle.

Yours also agrees better with the simulation. So I guess I made an error somewhere.

 

[sysprog]

There is no smallest number greater than zero.

I didn't mean to suggest that I thought there to be such a thing as a least number among the positive reals; mea culpa for any notional abuse on my part; I am not at odds with the fundamental theorem of calculus, and I accept the epsilon-delta definition of limits, as described here: https://mathworld.wolfram.com/Epsilon-DeltaDefinition.html.

I was seeking to evoke an acknowledgment that at any finite time greater than zero, the leftward acceleration of falling $m$ is non-zero $-$ that despite its acceleration not being the same as that of the large block and pulley, it is just as true for it that it is non-zero at any arbitrarily small finite $t>0$, as it is for the large block and pulley.

I think that there is no incompatibility between this statement of yours:

So, there is no finite initial time during which the block does not move.

and this statement of mine:

I think that at any $t>0$ the leftward acceleration of falling $m$ is also greater than zero.

That seeems to me to be obscured in some aspects of the following exchange:
I posted:

Isn't the leftward acceleration of $m$ at that point also greater than zero? I think that at any $t>0$ the leftward acceleration of falling $m$ is also greater than zero. Do you agree with that?

and @Baluncore posted:

Isn't the leftward acceleration of $m$ at that point also greater than zero?

Yes. All the plots of acceleration, velocity and position pass through the origin at t = 0.
On release at t = 0, the big M block gains a fixed acceleration.
At t = 0, the falling m block has zero horizontal acceleration.

I think that "on release at t = 0" might be more clearly expressed as "immediately at $t>0$", and similarly, I think that the hanging $m$ block isn't rightly called "falling" until $t>0$. However, I think that the first sentence of @Baluncore's response is abundantly clear: he said "Yes" to the question.